# Rewrite the expression f(x) = \frac {x^2+8}{x^2+3x+2} in terms of partial fraction.

## Question:

Rewrite the expression {eq}f(x) = \frac {x^2+8}{x^2+3x+2} {/eq} in terms of partial fraction.

## Partial Fraction:

A rational function is defined as the function of the form {eq}\dfrac{p(x)}{q(x)} {/eq}, where {eq}q(x)\neq 0 {/eq}.

When the degree of the numerator is less than the degree of the denominator, then the fraction is said to be proper.

- The partial fraction conversion is stated below:-

{eq}\dfrac{px+q}{(x-a)(x-b)}=\dfrac{A}{x-a}+\dfrac{B}{x-b} \ , \ a\neq b {/eq}

## Answer and Explanation:

Given: {eq}f(x) = \dfrac{x^2+8}{x^2+3x+2} {/eq}

Factorizing the denominator, we get:-

{eq}f(x) = \dfrac {x^2+8}{x^2+2x+x+2}\\\Rightarrow f(x)=\dfrac{x^2+8}{(x+1)(x+2)}\\\Rightarrow f(x)=\dfrac{x^2}{(x+1)(x+2)}+\dfrac{8}{(x+1)(x+2)}\\\Rightarrow f(x)=\dfrac{x^2}{(x+1)(x+2)}+\dfrac{1}{x+1}-\dfrac{1}{x+2}\\\Rightarrow f(x)=\dfrac{x^2-1+1}{(x+1)(x+2)}+\dfrac{1}{x+1}-\dfrac{1}{x+2}\\\Rightarrow f(x)=\dfrac{(x+1)(x-1)}{(x+1)(x+2)}+\dfrac{2}{x+1}-\dfrac{2}{x+2}\\\Rightarrow \boxed{ f(x)=\dfrac{x-1}{x+2}+\dfrac{2}{x+1}-\dfrac{2}{x+2}} {/eq}

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from High School Algebra I: Help and Review

Chapter 3 / Lesson 26